A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 5, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 5, what is the area of the triangle?

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Answer 1 · 2016-01-22T10:19:07

Area= $8.53766$ $8.53766$ square units

Explanation:

From the given, two angles $A = \frac{π}{6}$ $A=pi/6$ , $C = \frac{7 π}{12}$ $C=(7pi)/12$ and included side $b = 5$ $b=5$ . Try drawing the triangle. See that angle $B = \frac{π}{4}$ $B=pi/4$ by computation using the formula $A + B + C = π$ $A+B+C=pi$ . Also , the altitude from angle C to side c can be called height $h$ $h$ is $h = b \cdot sin (\frac{π}{6}) = 2.5$ $h = b*sin (pi/6)=2.5$
Side $c$ $c$ can be computed using formula $c = b \cdot cos A + h \cdot cot B$ $c=b*cos A+h*cot B$ .
$c = 5 \cdot cos (\frac{π}{6}) + 2.5 \cdot cot (\frac{π}{4})$ $c=5*cos (pi/6)+2.5*cot (pi/4)$ = $2.5 \cdot (\sqrt{3} + 1)$ $2.5*(sqrt3+1)$

$c = 2.5 (\sqrt{3} + 1)$ $c=2.5(sqrt3+1)$
Area can now be computed

Area $= \frac{1}{2} \cdot b \cdot c \cdot sin A$ $=1/2*b*c*sin A$

Area $= \frac{1}{2} \cdot 5 \cdot (2.5 (\sqrt{3} + 1)) \cdot sin (\frac{π}{6})$ $=1/2*5*(2.5(sqrt3+1))*sin (pi/6)$

Area $= 8.53766$ $=8.53766$ square units

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 5, what is the area of the triangle?

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Explanation:

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 5, what is the area of the triangle?